Solve 3300=16,500(1+0.05)^x

$$3300 = 16,500 ( 1 + 0.05 ) ^ { x }$$

Solve for x

$$x=-\log_{1.05}\left(5\right)\approx -32.986933736$$

Steps Using Definition of Logarithm

Divide both sides by 16500.

$$\frac{3300}{16500}=\left(1+0.05\right)^{x}$$

Reduce the fraction \frac{3300}{16500} to lowest terms by extracting and canceling out 3300.

$$\frac{1}{5}=\left(1+0.05\right)^{x}$$

Add 1 and 0.05 to get 1.05.

$$\frac{1}{5}=1.05^{x}$$

Swap sides so that all variable terms are on the left hand side.

$$1.05^{x}=\frac{1}{5}$$

Take the logarithm of both sides of the equation.

$$\log(1.05^{x})=\log(\frac{1}{5})$$

The logarithm of a number raised to a power is the power times the logarithm of the number.

$$x\log(1.05)=\log(\frac{1}{5})$$

Divide both sides by \log(1.05).

$$x=\frac{\log(\frac{1}{5})}{\log(1.05)}$$

By the change-of-base formula \frac{\log(a)}{\log(b)}=\log_{b}\left(a\right).

$$x=\log_{1.05}\left(\frac{1}{5}\right)$$

Solve for x (complex solution)

$$x=\frac{i\times 2\pi n_{1}}{\ln(1.05)}-\log_{1.05}\left(5\right)$$

$$n_{1}\in \mathrm{Z}$$